Question

How do you factor completely `x^3 – 2x^2 – 4x – 8`?

Answer 1

`x^3-2x^2-4x-8 = (x-x_1)(x-x_2)(x-x_3)`

where `x_1 = 2/3(1+root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33)))`

and `x_2, x_3` are related Complex zeros...

Explanation:

`f(x) = x^3-2x^2-4x-8`

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Discriminant

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example, `a=1`, `b=-2`, `c=-4` and `d=-8`, so we find:

`Delta = 64+256-256-1728-1152 = -2816`

Since `Delta < 0` this cubic has `1` Real zero and `2` non-Real Complex zeros, which are Complex conjugates of one another.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

`0=27f(x)=27x^3-54x^2-108x-216`

`=(3x-2)^3-48(3x-2)-304`

`=t^3-48t-304`

where `t=(3x-2)`

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Cardano's method

We want to solve:

`t^3-48t-304=0`

Let `t=u+v`.

Then:

`u^3+v^3+3(uv-16)(u+v)-304=0`

Add the constraint `v=16/u` to eliminate the `(u+v)` term and get:

`u^3+4096/u^3-304=0`

Multiply through by `u^3` and rearrange slightly to get:

`(u^3)^2-304(u^3)+4096=0`

Use the quadratic formula to find:

`u^3=(304+-sqrt((-304)^2-4(1)(4096)))/(2*1)`

`=(304+-sqrt(92416-16384))/2`

`=(304+-sqrt(76032))/2`

`=(304+-48sqrt(33))/2`

`=152+-24sqrt(33)`

`=2^3*(19+-3(33))`

Since this is Real and the derivation is symmetric in `u` and `v`, we can use one of these roots for `u^3` and the other for `v^3` to find Real root:

`t_1=2root(3)(19+3sqrt(33))+2root(3)(19-3sqrt(33))`

and related Complex roots:

`t_2=2omega root(3)(19+3sqrt(33))+2omega^2 root(3)(19-3sqrt(33))`

`t_3=2omega^2 root(3)(19+3sqrt(33))+2omega root(3)(19-3sqrt(33))`

where `omega=-1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.

Now `x=1/3(2+t)=2/3(1+t/2)`. So the zeros of our original cubic are:

`x_1 = 2/3(1+root(3)(19+3sqrt(33))+root(3)(19-3sqrt(33)))`

`x_2 = 2/3(1+omega root(3)(19+3sqrt(33))+omega^2 root(3)(19-3sqrt(33)))`

`x_3 = 2/3(1+omega^2 root(3)(19+3sqrt(33))+omega root(3)(19-3sqrt(33)))`